Question

Answer the question ASAP​

Answer 1

Consider the given sum.

$\displaystyle\sum_{r=1}^{n}a_r=\frac{n(n+1)(n+2)}{3},\ \ \ \forall n\geq 1$

Let me write the RHS as a combination.

$\displaystyle\sum_{r=1}^{n}a_r=\ ^{n+2}C_3$

Now we remember the given below one. It always holds true.

$\large\boxed{\displaystyle\sum_{r=k}^{n}^r\!C_k\ =\ ^{n+1}C_{k+1}\ \ \ \text{where}\ \ k\in\mathbb{W},\ \ k\leq n}$

As examples, this just means,

$1.\ \ \ ^1C_1=\ ^2C_2\\ \\ 2.\ \ \ ^0C_0+\ ^1C_0+\ ^2C_0=\ ^3C_1\\ \\ 3.\ \ \ ^4C_4+\ ^5C_4+\ ^6C_4+\ ^7\!C_4+\ ^8C_4+\ ^9C_4=\ ^{10}C_5$

What's the proof for this? Just consider Pascal's triangle!

$\begin{tabular}{ccccccccccc}&&&&&1&&&&&\\ &&&&1&&1&&&&\\ &&&1&&2&&1&&&\\ &&1&&3&&3&&1&&\\ &1&&4&&6&&4&&1&\\ 1&&5&&10&&10&&5&&1\end{tabular}$

In terms of combinations, Pascal's triangle is actually,

$\begin{tabular}{ccccccccccc}&&&&&\ ^0C_0&&&&&\\ &&&&\ ^1C_0&&\ ^1C_1&&&&\\ &&&\ ^2C_0&&\ ^2C_1&&\ ^2C_2&&&\\ &&\ ^3C_0&&\ ^3C_1&&\ ^3C_2&&\ ^3C_3&&\\ &\ ^4C_0&&\ ^4C_1&&\ ^4C_2&&\ ^4C_3&&\ ^4C_4&\\ \ ^5C_0&&\ ^5C_1&&\ ^5C_2&&\ ^5C_3&&\ ^5C_4&&\ ^5C_5\end{tabular}$

We know that each term in this triangle is produced by summing up the two terms on top of either sides of that term.

And also we see that,  $^0C_0\ =\ ^1C_1\ =\ ^2C_2\ =\ ^3C_3\ =\ ^4C_4\ =\ \dots\dots\ =\ ^nC_n\ =\ 1$

So,

$\bullet\ \ ^2C_1\ =\ ^1\!C_0+\ ^1\!C_1\ =\ ^1\!C_0+\ ^0\!C_0\\ \\ \bullet\ \ ^3C_1\ =\ ^2C_0+\ ^2C_1\ =\ ^2C_0+\ ^1\!C_0+\ ^1C_1\ =\ ^2C_0+\ ^1\!C_0+\ ^0\!C_0\\ \\ \bullet\ \ ^{10}C_5\ =\ ^9C_4+\ ^9C_5\ =\ ^9C_4+\ ^8C_4+\ ^8C_5\\ \\ =\ ^9C_4+\ ^8C_4+\ ^7C_4+\ ^7C_5\ =\ ^9C_4+\ ^8C_4+\ ^7C_4+\ ^6C_4+\ ^6\!C_5\\ \\ =\ ^9C_4+\ ^8C_4+\ ^7C_4+\ ^6C_4+\ ^5C_4+\ ^5C_5\\ \\ =\ ^9C_4+\ ^8C_4+\ ^7C_4+\ ^6C_4+\ ^5C_4+\ ^4C_4$

Finally,

$\begin{aligned}^{n+1}\!C_{k+1}\ &=\ ^n\!C_k+\ ^n\!C_{k+1}\ =\ ^n\!C_k+\ ^{n-1}\!C_k+\ ^{n-1}C_{k+1}\\ \\ &=\ ^n\!C_k+\ ^{n-1}\!C_k+\ ^{n-2}C_k+\ ^{n-2}C_{k+1}\\ \\ &=\ ^n\!C_k+\ ^{n-1}\!C_k+\ ^{n-2}C_k+\ ^{n-3}C_k+\ ^{n-3}C_{k+1}\\ \\ &=\ \dots\dots\dots\ \\ \\ &=\ ^n\!C_k+\ ^{n-1}\!C_k+\ ^{n-2}C_k+\ ^{n-3}C_k+\ \dots\ +\ ^{k+1}C_k+\ ^kC_k\\ \\ &=\ \sum_{r=k}^{n}^r\!C_k\end{aligned}$

So, come to our question.

$\displaystyle\sum_{r=1}^{n}a_r\ =\ \frac{n(n+1)(n+2)}{6}\ =\ ^{n+2}C_3$

According to our concept, we get that  $a_n\ =\ ^{n+1}\!C_2$

So,

$\displaystyle\lim_{n\to \infty}\ \sum_{r=1}^n\frac{1}{a_r}\ =\ \lim_{n\to \infty}\ \sum_{r=1}^{n}\dfrac{1}{^{r+1}C_2}\ =\ \lim_{n\to\infty}\ \sum_{r=1}^n\dfrac{2!(r-1)!}{(r+1)!}\\ \\ \\ =\ \lim_{n\to\infty}\ \sum_{r=1}^n\frac{2}{r(r+1)}\ =\ \lim_{n\to\infty}\ 2\sum_{r=1}^n\frac{1}{r(r+1)}\\ \\ \\ =\ 2\lim_{n\to\infty}\ \sum_{r=1}^n\frac{1}{r(r+1)}$

Now, we should be familiar with this too.

$\displaystyle\sum_{r=1}^n\frac{1}{r(r+1)}\ =\ \dfrac{n}{n+1}$

This can easily be proved by principle of mathematical induction.

We found P(1) true. So on assuming P(k) true and considering P(k + 1),

$\displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r+1)}\ =\ \left(\sum_{r=1}^k\frac{1}{r(r+1)}\right)+\dfrac{1}{(k+1)(k+2)}\\ \\ \\ =\ \dfrac{k}{k+1}+\dfrac{1}{(k+1)(k+2)}\\ \\ \\ =\ \dfrac{1}{k+1}\left(k+\dfrac{1}{k+2}\right)\\ \\ \\ =\ \dfrac{1}{k+1}\cdot\dfrac{(k+1)^2}{k+2}\ =\ \dfrac{k+1}{k+2}$

So,

$\displaystyle2\lim_{n\to\infty}\ \sum_{r=1}^n\dfrac{1}{r(r+1)}\ =\ 2\lim_{n\to\infty}\ \frac{n}{n+1}$

We know a fraction having denominator greater than numerator by 1 always tends to 1.

$n\longrightarrow \infty \ \ \implies\ \ \dfrac{n}{n+1}\longrightarrow 1$

So,

$\displaystyle2\lim_{n\to\infty}\ \frac{n}{n+1}\ =\ 2\times 1=\mathbf{2}$

So the answer is 2.

Hence option (3) is the answer.

Answer 2

Solution :-

We are provided with :-

$\displaystyle{{\sum\limits_{r=1}^{n}}a_r = \dfrac{n(n+1)(n+2)}{6}}$

Now first we will find out the term for nth term for this series.

S(n) - S(n-1) = nth term

$= \dfrac{n(n+1)(n+2)}{6} - \dfrac{(n-1)(n)(n+1)}{6}$

$= \dfrac{n}{6}\left( (n+1)(n+2) - (n-1)(n+1) \right)$

$= \dfrac{n}{6} \left( n^2 + 3n + 2 - n^2 + 1 \right)$

$= \dfrac{n}{6}(3n + 3)$

$= \dfrac{3.n.(n+1)}{6}$

$= \dfrac{n.(n+1)}{2}$

So our Equation for rth term

$= \dfrac{r.(r+1)}{2}$

Now

$\displaystyle{\lim_{n \to \infty}\ {\sum\limits_{r=1}^{n}} \dfrac{1}{a_r} = \lim_{n \to \infty} {\sum\limits_{r=1}^{n}} \dfrac{2}{r(r+1)}}$

$\displaystyle{= 2 \lim_{n \to \infty}\ {\sum\limits_{r=1}^{n}} \dfrac{1}{r(r+1)}}$

$\displaystyle{= 2 \lim_{n \to \infty} {\sum\limits_{r=1}^{n}} \dfrac{(r+1)- r}{r(r+1)}}$

$\displaystyle{ = 2 \lim_{n \to \infty} {\sum\limits_{r=1}^{n}} \dfrac{1}{r} - \dfrac{1}{r+1}}$

$\displaystyle{= 2 \lim_{n \to \infty} \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} ....... \dfrac{1}{n} - \dfrac{1}{n+1}}$

$\displaystyle{= 2 \lim_{n \to \infty} \dfrac{1}{1} - \dfrac{1}{n+1}}$

$\displaystyle{ = 2 \lim_{n \to \infty} \dfrac{(n+1) - 1}{n+1}}$

$\displaystyle{= 2 \lim_{n \to \infty} \dfrac{n}{n+1}}$

$\displaystyle{ = 2 \dfrac{\infty}{\infty+1}}$

Now as the number obtained will be just a very very little less than 1 ,, that is the number will tends to become 1 .

$= 2 \times 1$

$= 2$